Let S (1,2,...,n). A permutation a,a...,, of S is said to be alternating if ₂ <₂><>₂< Example: n Alternating Permutations 11 Example: 2 12 3 4 5 132, 231 2 1324, 1423, 2314, 2413, 3412 5 13254, 14253, 14352, 15243, 16 15342, 23154, 24153,24351, 25143, 25341, 34152,34251, 35142, 35241, 45132, 45231 A permutation apa₂...a of S is said to be reverse alternating if a><><₂>.... No. of such permutations 1 1 n| Reverse Alternating Permutations | No. of such permutations 11 1 2 21 3 312, 213 4 5 4231, 4132, 3241,3142,2143 53412, 52413, 52314, 51423, 51324, 43512, 42513, 42315, 41523, 41325, 32514, 32415, 31524, 31425, 21534, 21435 1 2 5 16 Use the Bijection Principle, to show that the number of alternating permutations of S is equal to the number of reverse alternating permutations of S.
Let S (1,2,...,n). A permutation a,a...,, of S is said to be alternating if ₂ <₂><>₂< Example: n Alternating Permutations 11 Example: 2 12 3 4 5 132, 231 2 1324, 1423, 2314, 2413, 3412 5 13254, 14253, 14352, 15243, 16 15342, 23154, 24153,24351, 25143, 25341, 34152,34251, 35142, 35241, 45132, 45231 A permutation apa₂...a of S is said to be reverse alternating if a><><₂>.... No. of such permutations 1 1 n| Reverse Alternating Permutations | No. of such permutations 11 1 2 21 3 312, 213 4 5 4231, 4132, 3241,3142,2143 53412, 52413, 52314, 51423, 51324, 43512, 42513, 42315, 41523, 41325, 32514, 32415, 31524, 31425, 21534, 21435 1 2 5 16 Use the Bijection Principle, to show that the number of alternating permutations of S is equal to the number of reverse alternating permutations of S.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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